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My name is Ann [436]
2 years ago
11

A developer wants to build a sprawling two-story office complex. The developer's and architect's vision is something low, modern

, and new. They are planning the complex in an urban area of town, replacing ten blocks of old high-rise buildings. Unfortunately, the zoning commission's vision does not coincide with either the developer's or the architect's. Which type of zoning regulation applies?
a. Directive Zoningb. Aesthetic Zoningc. Bulk Zoningd. Both A and B
Physics
1 answer:
Lera25 [3.4K]2 years ago
4 0

Answer:

a. directive zoning

Explanation:

Directive zoning is an instrument used in master plans, whereby the city is divided into areas on which differentiated land use and land use guidelines apply, especially urban indexes. Directive zoning acts primarily by controlling two main elements: the use and size (or size) of lots and buildings. It is therefore assumed that the end result achieved through individual actions is in line with the municipality's objectives, which include proportionality between occupation and infrastructure, the need to protect fragile areas and / or cultural interest, the harmony from the volumetric point of view, etc.

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A container contains 200g of water at initial temperature of 30°C. An iron nail of mass 200g at temperature of 50°C is immersed
andriy [413]

Answer:

The final temperature is 31.94°

Explanation:

The mass of the water in the container m₁ = 200 g = 0.2 kg

The initial temperature of the water,  T₁₁ = 30°C

The mass of the iron, m₂ = 200 g = 0.2 kg

The temperature of the iron T₂₁= 50°C is immersed in the water,

The specific heat capacity of the water, c₁ = 4200 J/(kg·°C)

The specific heat capacity of the iron, c₂ = 450 J/(kg·°C)

Heat capacity relation is given by the formula;

Heat capacity Q = Mass, m × Specific heat capacity, c × Temperature change, (T₂ - T₁)

Given that energy can neither be created nor destroyed, and with the assumption that all the heat lost by the nail is gained by the water we have;

Heat lost by iron nail = Heat gained by the  water

m₁ × c₁ × (T₂ - T₁₁) = m₂ × c₂ × (T₂₁ - T₂)

Where, T₂ is the final temperature

0.2 kg × 4200 J/(kg·°C) × (T₂ - 30) = 0.2 kg × 450 J/(kg·°C) × (50° - T₂)

840·T₂ - 25200 = 4500 - 90·T₂

4500 + 25200 = 840·T₂ + 90·T₂

29700 = 930·T₂

T₂ = 29700/930 = 31.94°.

The final temperature = 31.94°.

4 0
2 years ago
An apple falls from an apple tree growing on a 20° slope. The apple hits the ground with an impact velocity of 16.2 m/s straight
EleoNora [17]

Apple hits the surface with speed 16.2 m/s

The angle made by the apple velocity with normal to the incline surface is given as 20 degree

now the component of velocity which is parallel to the surface and perpendicular to the surface is given as

v_{perpendicular} = v cos20

v_{parallel} = v sin20

so here we have

v_{parallel} = 16.2 sin20

v_{parallel} = 5.5 m/s

<em>so its velocity along the incline plane will be 5.5 m/s</em>

7 0
2 years ago
A charge of 5.67 x 10-18 C is placed 3.5 x 10 m away from another charge of - 3.79 x 10 "C
miskamm [114]

Answer:

1. 579 x 10 ^-22N

Explanation:

F = kq1q2/r^2

   = 9.0 x 10^9 x 5.67 x 10^-18 x 3.79 x 10^-18/ (3.5 x 10^-2)^2

    = 1. 579 x 10 ^-22N

6 0
2 years ago
A long, straight wire carrying a current of 3.45 A moves with a constant speed v to the right. A 5-turn circular coil of diamete
d1i1m1o1n [39]

Answer:

I = 69.3  μA

Explanation:

Current through the straight wire, I = 3.45 A

Number of turns, N = 5 turns

Diameter of the coil, D = 1.25 cm

Resistance of the coil, R = 3.25 \mu ohms

Distance of the wire from the center of the coil, d = 20 cm = 0.2 m

The magnetic field, B₁, when the wire is at a distance, d, from the center of the coil.

B_{1} = \frac{\mu_{0}I }{2\pi d}

B_{1} = \frac{4\pi* 10^{-7}  *3.45 }{2\pi *0.2}\\B_{1} =0.00000345 T

Magnetic field B₂ when the wire is at a distance, 2d from the center of the coil

B_{2} = \frac{\mu_{0}I }{2\pi(2d)) } \\B_{2} = \frac{\mu_{0}I }{4\pi d } \\

B_{2} = \frac{4\pi* 10^{-7}  *3.45 }{2\pi *2*0.2}\\B_{2} = 0.000001725 T

Change in the magnetic field, ΔB = B₂ - B₁ = 0.00001725 - 0.0000345

ΔB = -0.000001725

Induced current, I = \frac{E}{R}

E = -N (Δ∅)/Δt

Δ∅ = A ΔB

Area, A = πr²

diameter, d = 0.0125 m

Radius, r = 0.00625 m

A = π* 0.00625²

A = 0.0001227 m²

Δ∅ =  -0.000001725 * 0.0001227

Δ∅ = -211.6575 * 10⁻¹²

E = -N (Δ∅)/Δt

E = -5\frac{-211.6575 * 10^{-12} }{4.70} \\E = 225.17 * 10^{-12} V

Resistance, R = 3.25 μ ohms = 3.25 * 10⁻⁶ ohms

I = E/R

I = \frac{225.17 * 10^{-12} }{3.25 * 10^{-6} }

I = 0.0000693 A

I = 69 .3 * 10⁻⁶A

I = 69.3  μA

3 0
2 years ago
You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By
Fittoniya [83]

Answer:

The frequency is    f  = 0.221 \ Hz

Explanation:

From the question we are told that  

     The  time taken for it to decay to half its original size is t  =  3.40 \ ms  =  3.40 *10^{-3} \ s

Let the voltage of the capacitor when it is fully charged be  V_o

Then the voltage of the capacitor at time t is  said to be  V  =  \frac{V_o}{2}

   Now  this voltage can be  mathematical represented as

      V  =  V_o  * e ^{-\frac{t}{RC} }

Where  RC  is the time constant

   substituting values  

    \frac{V_o}{2}  =  V_o  *  e ^{-\frac{3.40 *10^{-3}}{RC} }

    0.5  =  e^{-\frac{3.40 *10^{-3}}{RC} }

    - \frac{0.5}{RC}  =  ln (0.5)

     -\frac{0.5}{RC} =  -0.6931

     RC  =  0.721

Generally the cross-over frequency for a low pass filter is mathematically represented as

          f  = \frac{1}{2 \pi  * RC  }

substituting values  

           f  = \frac{1}{2*  3.142  * 0.72  }

           f  = 0.221 \ Hz

7 0
2 years ago
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